Question

circle the correct solution for questions 1 - 3.

1. which of the following best represents the domain of the function?

a. all real numbers
b. $-6 < x < 6$
c. $-2 \leq y \leq 5$
d. $-6 < x \leq 6$

1. which of the following best represents the range of the function?

a. all real numbers
b. $y \leq 3$
c. $y = 3$
d. $x \leq 3$

1. mrs. alvarez asked her students to create a graph with a domain and range of all real numbers. which student completed the task correctly?

a. tim only
b. travis only
c. both tim and travis
d. neither tim nor travis

Explanation:

Question 1

Step1: Analyze the domain of a function

The domain of a function is the set of all possible \(x -\)values. Looking at the graph, the \(x -\)values start just after \(x=-6\) (open - circle) and end at \(x = 6\) (closed - circle). So the domain is \(-6\lt x\leq6\).

Question 2

Step1: Analyze the range of a function

The range of a function is the set of all possible \(y -\)values. Looking at the second graph, the \(y -\)value is always \(y = 3\) (a horizontal line at \(y = 3\)).

Question 3

Step1: Analyze the domain and range of Tim's graph

For Tim's graph (a V - shaped graph), as \(x\) can take any real value (extends infinitely to the left and right), and \(y\) can also take any non - negative real value (but if we consider the full - line behavior, for a linear - like V - shape (assuming it's a linear transformation), if we extend it infinitely in both \(x\) directions, \(y\) can take all real numbers (if we consider the general form \(y=|x|\) shifted or transformed in a non - restricted way for \(y\) in the context of domain and range as all real numbers).

Step2: Analyze the domain and range of Travis's graph

For Travis's graph (a straight line), since a straight line (assuming it's of the form \(y=mx + b\) with \(m
eq0\)) has a domain of all real numbers (\(x\) can be any real number) and a range of all real numbers (\(y\) can be any real number as \(x\) varies over \(\mathbb{R}\)).

Answer:

1. d. \(-6\lt x\leq6\)
2. c. \(y = 3\)
3. c. Both Tim and Travis